In the arrangement shown in the figure,when t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $V_0$ be the potential difference across the potentiometer wire $AB$ of length $L$. The potential gradient is $k = V_0/L$.$CASE$ $1$: When swi

Vedclass · Accepted Answer

$2\,\Omega, 12\,V$

Vedclass · Answer

(B) Let $V_0$ be the potential difference across the potentiometer wire $AB$ of length $L$. The potential gradient is $k = V_0/L$.$CASE$ $1$: When switch $S_2$ is open,the galvanometer shows no deflection at $l = L/2$. This means the potential difference across the length $l$ is equal to the $emf$ of the $6\, V$ cell.$k \cdot (L/2) = 6\, V$$(V_0/L) \cdot (L/2) = 6\, V \Rightarrow V_0/2 = 6\, V \Rightarrow V_0 = 12\, V$.$CASE$ $2$: When switch $S_2$ is closed,the galvanometer shows no deflection at $l = 5L/12$. The potential difference across the length $l = 5L/12$ is equal to the terminal voltage of the $6\, V$ cell.Terminal voltage $V_t = k \cdot (5L/12) = (V_0/L) \cdot (5L/12) = (12/L) \cdot (5L/12) = 5\, V$.When $S_2$ is closed,the $6\, V$ cell is connected to a $10\,\Omega$ external resistor. The terminal voltage is given by $V_t = E_{cell} - I r$,where $I = E_{cell} / (R + r)$.$V_t = E_{cell} \cdot [R / (R + r)]$$5 = 6 \cdot [10 / (10 + r)]$$5(10 + r) = 60$$50 + 5r = 60$$5r = 10 \Rightarrow r = 2\,\Omega$.Thus,$r = 2\,\Omega$ and $E = 12\, V$.

Similar Questions

The $e.m.f.$ of a standard cell balances across $150 \ cm$ length of a potentiometer wire. When a resistance of $2 \ \Omega$ is connected as a shunt across the cell,the balance point is obtained at $100 \ cm$. The internal resistance of the cell is .............. $\Omega$.

In a potentiometer,a balance point is obtained when:

The resistance of $10\, m$ long potentiometer wire is $1\,\Omega/m$. $A$ cell of $e.m.f.$ $2.2\, V$ and a high resistance box are connected in series to this wire. The value of resistance taken from the resistance box for getting a potential gradient of $2.2\, mV/m$ will be ............... $\Omega$.

$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module